Then \(f(z_0) = 0\)  and \(f^{\prime }(z_0) = \sqrt{\pi ^{-1}v(E)}\) . The Ahlfors-Beurling inequality states that \(|f(z)| \le 1\)  for all \(z \in \mathbb {C}_{\infty }\) . By the maximum modulus theorem, \(|f(z)| < 1\)  on \(\Omega \) . Thus, \(f\)  is an extremal function for \(c_B\) , which implies \(\sup _{z \in \Omega } |f(z)| = 1\) . By continuity there exists \(z_2 \in \partial \Omega \)  such that \(|f(z_2)| = 1\) . We observe, after a careful inspection of the proof of the Ahlfors-Beurling inequality, as given in [1]}, that \(E\)  must be a union of a closed disk with a closed measure-zero set. For completeness we resupply the proof since this observation is not stated in the literature, cf. [2]}, [3]}, [1]}. Indeed, after a rotation and translation of \(E\)  we may assume that
\(1 = |f(z_2)| = {1 \over \sqrt{\pi v(E)}}\int _{E} {1 \over w} dv(w).\) 
